Compute the values of the product

1 + 
1
1


1 + 
1
2


1 + 
1
3

    

1 + 
1
n

 
for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture.

1 + 
1
1


1 + 
1
2


1 + 
1
3

    

1 + 
1
n

 = 
  
Correct: Your answer is correct.
Prove your conjecture by mathematical induction.
Proof (by mathematical induction): Let P(n) be the equation 

1 + 
1
1


1 + 
1
2


1 + 
1
3

    

1 + 
1
n

 = 
  

Let's start by computing the values of the product for small values of \( n \). 

The product in question is:

\[
P(n) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \cdots \left(1 + \frac{1}{n}\right)
\]

### Step 1: Compute for small values of \( n \).

- For \( n = 1 \):
  \[
  P(1) = 1 + \frac{1}{1} = 2
  \]

- For \( n = 2 \):
  \[
  P(2) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) = 2 \times \frac{3}{2} = 3
  \]

- For \( n = 3 \):
  \[
  P(3) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) = 2 \times \frac{3}{2} \times \frac{4}{3} = 4
  \]

- For \( n = 4 \):
  \[
  P(4) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \left(1 + \frac{1}{4}\right) = 2 \times \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} = 5
  \]

- For \( n = 5 \):
  \[
  P(5) = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \left(1 + \frac{1}{4}\right) \left(1 + \frac{1}{5}\right) = 2 \times \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} = 6
  \]

From the pattern, we can conjecture that the general form of the product is:

\[
P(n) = n + 1
\]

### Step 2: Prove by mathematical induction.

We will prove that \( P(n) = n + 1 \) by induction.

#### **Base Case**:
For \( n = 1 \):
\[
P(1) = 2 = 1 + 1
\]
The base case holds true.

#### **Inductive Step**:
Assume that for some \( k \geq 1 \), the formula holds, i.e., 
\[
P(k) = k + 1
\]
We need to prove that the formula holds for \( k + 1 \), i.e., 
\[
P(k+1) = (k+1) + 1 = k + 2
\]

Using the inductive hypothesis:
\[
P(k+1) = P(k) \times \left(1 + \frac{1}{k+1}\right)
\]
Substitute the inductive hypothesis \( P(k) = k + 1 \):
\[
P(k+1) = (k + 1) \times \left(1 + \frac{1}{k+1}\right) = (k + 1) \times \frac{k+2}{k+1} = k + 2
\]

Thus, the formula holds for \( k + 1 \), completing the inductive step.

By mathematical induction, the formula \( P(n) = n + 1 \) is true for all \( n \geq 1 \).

---

Would you like more details or have any questions about this process?

Here are 5 relative questions for further exploration:
1. How does the product change if the terms were \( 1 - \frac{1}{n} \) instead of \( 1 + \frac{1}{n} \)?
2. What happens if the sum instead of the product is taken for the same terms?
3. How can this result be generalized for non-integer values of \( n \)?
4. Can this formula be extended to complex numbers?
5. What role does the harmonic series play in this type of product?

**Tip**: Mathematical induction is a powerful technique, often used to prove statements for sequences or recursive relations. Familiarize yourself with its base case and inductive step structure.

Compute the values of the product 1 + 1/1, 1 + 1/2, 1 + 1/3, ..., 1 + 1/n for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture. Prove your conjecture by mathematical induction.

Discover how to compute the product of the series 1 + 1/1, 1 + 1/2, ..., 1 + 1/n and conjecture a general formula. This lesson includes a full proof using mathematical induction, perfect for students in Grades 10-12.

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